Question: Simplify and expand the following expression: $ \dfrac{z}{4z - 8}-\dfrac{-10}{z + 10} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4z - 8)(z + 10)$ Multiply the first term by $\dfrac{z + 10}{z + 10}$ $ \begin{align*} \dfrac{z}{4z - 8} \times \dfrac{z + 10}{z + 10} & = \dfrac{(z)(z + 10)}{(4z - 8)(z + 10)} \\ & = \dfrac{z^2 + 10z}{(4z - 8)(z + 10)}\end{align*} $ Multiply the second term by $\dfrac{4z - 8}{4z - 8}$ $ \begin{align*} \dfrac{-10}{z + 10} \times \dfrac{4z - 8}{4z - 8} & = \dfrac{(-10)(4z - 8)}{(z + 10)(4z - 8)} \\ & = \dfrac{-40z + 80}{(z + 10)(4z - 8)}\end{align*} $ Now we have: $ = \dfrac{z^2 + 10z}{(4z - 8)(z + 10)} - \dfrac{-40z + 80}{(z + 10)(4z - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{z^2 + 10z - (-40z + 80)}{(4z - 8)(z + 10)} $ $ = \dfrac{z^2 + 10z + 40z - 80}{(4z - 8)(z + 10)} $ $ = \dfrac{z^2 + 50z - 80}{(4z - 8)(z + 10)}$ Expand the denominator: $ = \dfrac{z^2 + 50z - 80}{4z^2 + 32z - 80}$